Chebyshev Inequality¶

Theorem¶

For $X$ with finite variance $\sigma^2$ , mean $\mu$ , and any $a>0$ $0$ ,

\mathbb{P}(|X-\mu|\ge a)\le\frac{\sigma^2}{a^2}.

Bounds can be capped at $1$ . If $\sigma=0$ $0$ , then $X=\mu$ almost surely and the deviation probability is zero.

The Inequality bounds a tail probability from above; it is not an equality or an approximation. Applying the Markov inequality to the nonnegative squared deviation $(X-\mu)^2$ gives this bound.

Interpretation¶

Finite variance guarantees a bound on how often a variable can be far from its mean, regardless of the shape of the distribution. Measured in standard deviations, at most $1/k^2$ of the probability lies $k\sigma$ or more from $\mu$ : at most $1/4$ beyond $2\sigma$ and at most $1/9$ beyond $3\sigma$ .

A bell-shaped density with both tails beyond mu minus a and mu plus a shaded; together the shaded tails have probability at most sigma squared over a squared.

The figure shades both tails beyond $\mu\pm a$ of a normal density with $a=2\sigma$ $2 σ$ . Chebyshev's inequality bounds the combined shaded probability by $\sigma^2/a^2=1/4$ $1/4$ , while for the normal distribution it is approximately $0.046$ . The bound is loose here because it must hold for every distribution with the same variance.

The proof applies the Markov inequality to $(X-\mu)^2$ $μ)^{2}$ , so only the variance enters: no symmetry, single peak or density is assumed. The bound is informative only for $a>\sigma$ $σ$ ; for $a\le\sigma$ the right-hand side is at least $1$ . Although it is loose for the normal distribution, it cannot be improved in general: if $a\ge\sigma$ and $X$ takes the values $\mu-a$ and $\mu+a$ with probability $\sigma^2/(2a^2)$ each and $\mu$ otherwise, then $X$ has variance $\sigma^2$ and $\mathbb{P}(|X-\mu|\ge a)=\sigma^2/a^2$ exactly.

Its main use is the weak law of large numbers. The sample mean of $n$ i.i.d. observations has variance $\sigma^2/n$ , so $\mathbb{P}(|\bar X_n-\mu|\ge\varepsilon)\le\sigma^2/(n\varepsilon^2)$ $σ^{2} / (n ε^{2})$ , which tends to $0$ for every fixed $\varepsilon>0$ $0$ : this is convergence in probability of $\bar X_n$ to $\mu$ .

Key Results¶

If $\sigma>0$ and $k>0$ $0$ , $\mathbb{P}(|X-\mu|\ge k\sigma)\le1/k^2$ $1/ k^{2}$ .
Applying the Markov inequality to $(X-\mu)^2$ with threshold $a^2$ proves the bound $\sigma^2/a^2$ .

Examples¶

For mean $100$ , standard deviation $5$ , and deviation threshold $15$ , the probability is at most $25/225=1/9$ $1/9$ .
For an i.i.d. sample with variance $4$ , $\mathbb{P}(|\bar X_n-\mu|\ge0.5)\le16/n$ $16/ n$ , giving $0.04$ at $n=400$ $400$ .

Prerequisites¶

Sources¶

Nicky van Foreest · Probability Distributions · Week 7

Aliases: Chebyshev's inequality, Bienaymé-Chebyshev inequality, Tchebysheff's inequality, Chebyshev's bound Symbols: epsilon, mu, sigma

Last updated: 20 October 2026 03:02